TS EAMCET · Maths · Functions
If \(f:[-6,6] \rightarrow R\) is defined by \(f(x)=x^2-3\) for \(x \in R\), then \[ (f \circ f \circ f)(-1)+(f \circ f \circ f)(0)+(f \circ f \circ f)(1) \] is equal to
- A \(f(4 \sqrt{2})\)
- B \(f(3 \sqrt{2})\)
- C \(f(2 \sqrt{2})\)
- D \(f(\sqrt{2})\)
Answer & Solution
Correct Answer
(A) \(f(4 \sqrt{2})\)
Step-by-step Solution
Detailed explanation
Given, \(\quad f(x)=x^2-3\) Now, \(\quad f(-1)=(-1)^2-3=-2\) \[ \begin{aligned} & \Rightarrow \quad f \circ f(-1)=f(-2)=(-2)^2-3=1 \\ & \Rightarrow \quad f \circ f \circ f(-1)=f(1)=1^2-3=-2 \end{aligned} \] Now, \[ f(0)=0^2-3=-3 \]…
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